#include <iostream>
using namespace std;
const int N = 210;
int f[N][N];
int n, k;
// f[i][j]表示总和为i，并且数的个数是j的所有方案数

int main()
{
    f[1][1] = 1;
    cin >> n >> k;
    for (int i = 2; i <= n; ++i)
        for (int j = 1; j <= i; ++j)
            f[i][j] = f[i - 1][j - 1] + f[i - j][j];
    cout << f[n][k] << endl;
    return 0;
}